I am solving the PDE $$\bigtriangleup u = 0,\ u(1,\ \theta) = \sin^2\theta$$
I plugged $u = R(r)T(\theta)$ into the polar Laplacian formula, and ended up with $\frac{r^2}{R}\frac{\partial^2 R}{\partial r^2} + \frac{r}{R}\frac{\partial R}{\partial r} = -\frac{1}{T}\frac{\partial^2 T}{\partial \theta^2} = \lambda$. My solutions to the Sturm-Louiville problems are $R = Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}$ and $T = Ce^{-\sqrt{-\lambda}\theta} + De^{\sqrt{-\lambda}\theta}$ (I have ruled out the $\lambda = 0$ case). Thus I have $u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(Ce^{-\sqrt\lambda\theta i} + De^{\sqrt\lambda\theta i})$, or by Euler's Formula $$u = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta))$$
I have heard several explanations for the next step, but the way it makes sense in my head (correct me if I'm wrong) is that we know the domain is a disk, because the given boundary condition describes a shape with constant radius which implies a circle, and a disk domain in turn implies that $u(r,\ \theta) = u(r,\ \theta + 2m\pi)$ for all $r$, $\theta$, and $m$, where $m$ is an arbitrary integer. This is like having an extra boundary condition for every $m$, and plugging into general solution yields $(Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta)) = (Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda})(C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi)))$
Since this must hold for all $r$, not just the special case where $(Ar^{-\sqrt\lambda} + Br^{\sqrt\lambda}) = 0$, I can divide by this factor to get $C\cos(\sqrt\lambda \theta) + D\sin(\sqrt\lambda \theta) = C\cos(\sqrt\lambda (\theta + 2m\pi)) + D\sin(\sqrt\lambda (\theta + 2m\pi))$, which has solutions $(C\ $arbitrary$, D\ $arbitrary$, \lambda = n^2)$, where $n$ is any integer, and $(C = 0, D = 0, \lambda\ $arbitrary$)$. The latter leads to $u = 0$ so let us proceed with $\lambda = n^2$, which gives $$u = \sum_{n = 0}^\infty (Ar^{-n} + Br^n)(C\cos(n \theta) + D\sin(n \theta))$$
Now applying the actual boundary condition produces $\sin^2\theta = \sum_{n = 0}^\infty (A + B)(C\cos(n \theta) + D\sin(n \theta))$, or $\sin^2\theta = \sum_{n = 0}^\infty E\cos(n \theta) + F\sin(n \theta)$ where $E = AC + BC$ and $F = AD + BD$. All Fourier Series terms are $0$ except at $n = 0$ and $n = 2$, which yield $E = \frac{1}{2} \implies C = \frac{1}{2A + 2B}$ and $E = -\frac{1}{2} \implies C = -\frac{1}{2A + 2B}$, respectively, and so $$u = \frac{(Ar^{-n} + Br^n)(1 – cos(2\theta))}{2A + 2B}$$
I am stuck here. I am out of boundary conditions and still have two arbitrary constants in the answer. Even if I set $A = 0$ and only consider solutions consistent with the steady temperature physical application of this equation, $B$ remains, and I would prefer to solve the problem purely mathematically. How can I continue solving the problem from here?
Best Answer
I think there is a problem with your last expression of $u$. If I follow your calculations and if I'm not mistaken (correct me if so), at the end I get:
$$u(r,\theta)=\frac{1}{2}-\frac{\cos(2\theta)\left(A\frac{1}{r^2}+Br^2\right)}{2(A+B)}$$
From there, if you stick to the physical interpretation, it would be pretty weird to have an infinite temperature at $r=0$. Hence we can agree on the condition:
$$|u(0,\theta)|<+\infty$$
It is easy to see that this implies $A=0$ (else there is a singularity at $r=0$), which provides:
$$u(r,\theta)=\frac{1}{2}-\frac{\cos(2\theta)\left(Br^2\right)}{2B}$$
And finally $B$ cancels out !
$$u(r,\theta)=\frac{1}{2}\left(1-\cos(2\theta)r^2 \right)$$